When powering the circuit for the first time it started "ringing" (hearing low f. noise). After running some tests the regulator stopped to take current (The measured output voltage was 32V though). I am not a professional analog circuit designer so my design might have problems.

Some suggestions why the circuit doesn't work:

Compensation network design

I calculated the component values according to the datasheet, but I am not sure if those are right. Although I simulated the circuit with LTspice and it gave a proper transient analysis.

board layout

I am not familiar with Kelvin grounds etc. so there might be something to adjust.

However the connections are good since I soldered and double-checked them myself (Experienced with SMA).

If you have any suggestions or you can help me with the compensation network, please reply.

I tried to follow the suggestions as closely as possible, but having only a 2-sided PCB means that the routing options are limited. I have the FE package as shown in Fig 17. Bottom layer is solid ground plane and I tried to keep all the ground connections as short as possible. Thanks for the feedback.

I found the solution for this problem. The datasheet suggested a inductor with a equation Lmin = 12/(Fosc) which is of course absurd to use in my case. I calculated now that 800 µH inductor is needed for this regulator. So I'll change my regulator...

Take another look at figure 17 on page 31 of the datasheet:http://cds.linear.com/docs/en/datasheet/31152f.pdf
The area under the IC has copper pour areas on BOTH SIDES of the PCB, with vias denoted by the 7 black dots. These vias are critical, as they conduct the heat from the underside of the IC, through the vias, to the bottom side of the board, where the heat may be removed from the area by radiation, convection and conduction. Without those vias and a ground plane on the other side, the IC will rapidly overheat and self-destruct.

It looks like you are trying to use a very physically large inductor on your layout; why so huge? Your traces are too narrow and long because of that.
You should upload your .asc and .brd files (you ARE using Cadsoft Eagle, right? Or what are you using?)

Take another look at figure 17 on page 31 of the datasheet:http://cds.linear.com/docs/en/datasheet/31152f.pdf
The area under the IC has copper pour areas on BOTH SIDES of the PCB, with vias denoted by the 7 black dots. These vias are critical, as they conduct the heat from the underside of the IC, through the vias, to the bottom side of the board, where the heat may be removed from the area by radiation, convection and conduction. Without those vias and a ground plane on the other side, the IC will rapidly overheat and self-destruct.

It looks like you are trying to use a very physically large inductor on your layout; why so huge? Your traces are too narrow and long because of that.
You should upload your .asc and .brd files (you ARE using Cadsoft Eagle, right? Or what are you using?)

Click to expand...

Yeah that selection of huge inductor was a mistake of mine. The details are trivial and pointless.

I have copper pour areas in both sides as well, but only two vias since the drill size is an issue where my board is manufactured.

Now to the inductor. I have read the Fundamentals of Power Electronics (by R. Erickson). It suggests that my problem might be the discontinuous current mode. "Buck-boost converter with small load current must be kept in continuous current mode..." Do you agree on this? I calculated the inductor value again according to the book's formulas in worst case scenario so that the converter should never go to DCM and got 400 µH for minimum inductance (not 800 µH). Again this is just theory, but I talked to a professor in power electronics about this problem and he confirmed this. I didn't quite clearly understand the problems of DCM, before I started to design the circuit. (This is only a part of the circuit so my focus might have slipped to another parts as well).

So the problem could lie in the layout design (too narrow and long traces to the inductor) and the small inductance. Thank you again for the efforts.